1. Electrostatics Wenny Maulina

3. Electric charge Protons have positive charge Electrons have negative charge Opposite signs attract Similar signs repel Electric field – used to calculate force between charges

7. Where do charges come from? Matter is made up of atoms. + + + – – – + – Proton (positive charge) neutron (neutral) electron (negative charge) atom nucleus

8. If electrons = protons  neutral Where do charges come from? If electrons > protons  gaining electrons, negative charge If electrons < protons  losing electrons, positive charge

9. Electric Charge The unit of electric charge is the coulomb (C). The smallest amount of charge that can be added or removed from an object is the elementary charge, e = 1.6  10 -19 C. The charge of a proton is + e, an electron - e. The charge of an object, Q, is always a multiple of this elementary charge: Q = N e, where N is an integer.

10. Example Hitunglah muatan total dari 40 kg proton?

11. Example

12. Rubbing materials does NOT create electric charges. It just transfers electrons from one material to the other. Where do charges come from?

13. Type of materials Conductors – Materials that allow electrons to flow through them easily – Materials, such as metals Insulators – Materials that do NOT allow electrons to flow through them easily. – Materials, such as rubber and glass Semiconductors – Materials has an electrical conductivity value between a conductor and isolator – Materials, such as silicon and germanium

14. CHARGING A METAL SPHERE BY INDUCTION Charges are free to move in a conductor but are tightly bound in an insulator. The earth (“ground”) is a large conductor having many free charges.

15. Electric Force This gives the force on charged object 2 due to charged object 1 The direction of the force is either parallel or antiparallel to this unit vector depending upon the relative signs of the charges is a unit vector pointing from object 1 to object 2 As with all forces, the electric force is a Vector So we rewrite Coulomb’s Law as q2q2 q1q1

17. Superposition of Forces If there are more than two charged objects interacting with each other The net force on any one of the charged objects is The vector sum of the individual Coulomb forces on that charged object

18. Example Tiga muatan positif diletakkan pada koordinat (0,0), (4,0), dan (9,0) dari suatu sistem koordinat. Satuan sistem koordinat dalam meter. Besar muatan tersebut berturut-turut 3μC, 6μC, dan 9μC. Hitunglah gaya yang bekerja pada muatan 6μC!

19. Electric Field Like the electric force, the electric field is also a vector If there is an electric force acting on an object having a charge q o, then the electric field at that point is given by (with the sign of q 0 included)

20. Electric Field If we know the electric field, we can calculate the force on any charge: The direction of the force depends on the sign of the charge – in the direction of the field for a positive charge, opposite to it for a negative one.

22. Electric Field The electric field of a point charge can then be shown to be given by Earlier we saw that the force on a charged object is given by The term in parentheses remains the same if we change the charge on the object at the point in question The quantity in the parentheses can be thought of as the electric field at the point where the test object is placed

23. Electric Field As with the electric force, if there are several charged objects, the net electric field at a given point is given by the vector sum of the individual electric fields

24. We can use our notion of the gravitational field to form the concept of an ELECTRIC FIELD (E) Recall force between two masses: F = m g g is the gravitational field (9.8 m/sec 2 ) | F | = G | M m | / r 2 The force between two charges Q and q o is given by: F = q o E | F | = k | Q q o | / r 2

25. Coulomb’s Law: | F | = k | Q q o | / r 2 Rearranged: | F | = | q o [k Q/r 2 ] | Gives us: F = q o E where the electric field E is: | E | = | k Q / r 2 |

26. Forces on electron beam in a TV tube (CRT) F = Q E and F = m g (vector equations)

27. Example

28. Sebuah elektron ditembakkan ke dalam medan listrik homogen E= 2000 N/C dengan kecepatan awal V 0 = 10 6 m/s tegak lurus medan. (a) bandingkan gaya listrik yang bekerja pada elektron dengan gaya gravitasi yang bekerja padanya, (b) seberapa jauh elektron dibelokkan setelah menempuh jarak 1 cm pada arah sumbu x. E - V0V0 q

29. Exercise q2q2 q3q3 q1q1 5 cm

30. Exercise A CD B